Rod End Velocity (Optimization)

Since the rod is rigid, the velocities of the two endpoints must have the same component parallel to the rod. In other words $3v_{x_2} + 2v_{y_2} \; = \; \left(\begin{array}{c} 3 \\ 2\end{array}\right) \cdot \left(\begin{array}{c} v_{x_2} \\ v_{y_2}\end{array}\right) \; = \; \left(\begin{array}{c} 3 \\ 2\end{array}\right) \cdot \left(\begin{array}{c} v_{x_1} \\ v_{y_1}\end{array}\right) \; = \; 3v_{x_1} + 2v_{y_1} \; = \; 6$ But then, using the Cauchy-Schwarz Inequality, $36 \; = \; (3v_{x_2} + 2v_{y_2})^2 \; \le \; (3^2 + 2^2)(v_{x_2}^2 + v_{y_2}^2)$ and hence $v_{x_2}^2 + v_{y_2}^2 \ge \boxed{\tfrac{36}{13}}$ , with equality when $v_{x_2} = \tfrac{18}{13}$ , $v_{y_2} = \tfrac{12}{13}$ .

A similar approach - we have $\left((1-2t)-(4+v_{x_2}t)\right)^2+\left((3+6t)-(5+v_{y_2}t)\right)^2=13$ and thus $(2+v_{x_2})^2t+6(2+v_{x_2})+(6-v_{y_2})^2t-4(6-v_{y_2})=0.$ Substituting $t=0$ yields $6(2+v_{x_2})=4(6-v_{y_2}),$ and hence $v_{y_2}=\dfrac{6-3v_{x_2}}{2}.$ Thus $\dfrac{d}{dx}\left(v_{x_2}^2+v_{y_2}^2\right)=\dfrac{d}{dx}\left(v_{x_2}^2+\left(\dfrac{6-3v_{x_2}}{2}\right)^2\right)=0\\ \Longrightarrow v_{x_2}=\dfrac{18}{13}\\ \Longrightarrow \text{min}\left(v_{x_2}^2+v_{y_2}^2\right)=\dfrac{36}{13}\approx \boxed{2.769}.$